# Calculus conservation of momentum problem

• putongren
In summary, the conversation discusses a problem that uses calculus and the conservation of momentum principle. The solution involves integration and the rate at which balls hit a moving car. The conversation also clarifies the meaning of the limits of integration and the role of variables in the integrand. The conversation also touches on the concept of the Doppler shift and provides a hint for solving a related problem.
putongren
Hello All,

I am trying to understand the solution of a problem that uses calculus and the conservation of momentum principle.The problem is here: http://www.physics.harvard.edu/academics/undergrad/probweek/prob79.pdf and the answer is here: http://www.physics.harvard.edu/academics/undergrad/probweek/sol79.pdf

I have several questions concerning the solution, and I will ask one question at a time. My understanding of integration is weak. In the solution, we arrive at:
$$\int_{M}^{m} \frac{dm}{m} = \int_{0}^{v} \frac{dv}{u-v}$$. Why is it that we integrate between m and M and v and 0? What does integrating those terms physically mean? I think there is something missing in my understanding.

Last edited by a moderator:
The limits of integration represent values taken initially (lower limit) and some time t later (upper limit).

Initially, the mass is M (car only, no balls) and the velocity is 0. Some time later, these quantities are m and v. Or put another way, the mass is M when the velocity is 0, and the mass is m when the velocity is v.

p.s.:
By the way, whoever wrote that solution was sloppy by making variables in the integrand the same as the limits, which can be considered as constants for the purposes of integration. For example, it should read

$$\int_{M}^{m} \frac{dm'}{m'} = \int_{0}^{v} \frac{dv'}{u-v'}$$​

where the primes (') are used to distinguish the variables of integration.

Redbelly98,

so m = mass of the car M plus the culmination of balls thrown inside the car at a later time?

Last edited:
Yes.

(This is implied by equation (1) in the solution you linked to.)

To further clarify things, m and v are functions and dm is the mass of the ball and dv is the change in the velocity of the car after being hit by the ball.

Another question, why is $$dm/dt$$ = $$(u-v)\sigma/u$$? I thought $$dm/dt$$ is the rate at which the ball hits the car, which is defined as $$\sigma$$.

σ is the rate at which the balls leave you (the person throwing them).

The rate at which the balls hit the moving car is different. For example, if the car moves at the same speed as the balls (v=u), the balls do not hit the car and the rate of mass increase must be zero in that case.

ok thanks for clarifying that issue. Sorry I'm at little slow, why is $$dm/dt$$ = $$(u-v)\sigma/u$$? Also, not sure how eqn (3) leads to (4), even though it is derived in the solution.

putongren said:
ok thanks for clarifying that issue. Sorry I'm at little slow, why is $$dm/dt$$ = $$(u-v)\sigma/u$$?
One would have to work out the rate at which mass arrives at the car, given u, v, and σ. If you assumed a single mass equal to σ·1s were thrown every second, with a speed of u, you could work out how often the masses would hit a car traveling at speed v (assumed constant for this calculation).

Also, not sure how eqn (3) leads to (4), even though it is derived in the solution.
Start with eqn. (2), and differentiate w.r.t. t. That gives an expression for dm/dt, which may be equated with the R.H.S. of (3). Separate the v and t parts of the equation, and integrate.

I understand how (2) leads to (3) leads to (4). However, I'm stuck still on the rate at which the balls hit car. It didn't even occur to me that the balls hitting the car would be different from balls leaving the thrower until you told me.

Yes, that is probably a common error when looking at problems like this. If you are familiar with the Doppler shift already, this is an analogous effect.

Suppose the car is traveling at speed v. When it is a distance L from the thrower, a ball of mass σ·Δt is thrown toward the car at velocity u.

Question #1: at what time t1 does this ball hit the car?

Next: at a time Δt later, another ball of mass σ·Δt is thrown at the car.
(Note: throwing a mass of σ·Δt at intervals spaced by Δt is consistent with the a mass-per-time rate of σ)
When the second ball is thrown, the car is at a distance L+v·Δt from the thrower.

Question #2: at what time t2 does this second ball hit the thrower?

Question #3: what is t2 - t1?

Note that a mass of σ·Δt hits the car every t2-t1 time interval, so the rate at which mass hits the car is

σ·Δt / (t2-t1)​

Use the answer to Question #3 to replace (t2-t1) in this expression, and find the rate at which mass hits the car.

Sorry I can't get t2. Tried really hard.

Okay, here's a starter-hint:

The second ball is thrown at time Δt. At this point, the car is a distance L+v·Δt from the thrower.

How long does this baseball take to reach the car? Use the same method you used to find t1, using the new starting distance L+v·Δt instead of just L.

Then add that time to Δt to get t2.

I got a ridiculous answer, but I will share it with you anyway.

Since distance = speed x time, L = initial distance between car and thrower, u = speed of ball, v1 = speed of car before being hit, v2 speed of car after being hit, t1 = time it takes for the first ball to hit, t2 = time it takes for the second ball to hit, Δt = time between first and second balls thrown.

$$\because$$ t1(u-v1) = L+v1t1
$$\therefore$$t1 = $$L/(u-2v)$$

The distance it takes the ball to catch up with the car is equal to the initial distance L plus how much the car travels during the time it takes the ball to hit the car.

(t1-Δt)(u-v1)+(t2-t1)(u-v2) = L+vt1+(t2-t1)v2
This is the equation of distance that the car travels before being hit by the second ball. The speed is different after the car being hit by the first ball since it has an additional mass.

Last edited:
Hi,

I haven't forgotten this, just haven't had time to make a good reply. Will try to respond tomorrow (Saturday).

That's not a bad start, I caught a couple of errors that I'll point out:
putongren said:
I got a ridiculous answer, but I will share it with you anyway.

Since distance = speed x time, L = initial distance between car and thrower, u = speed of ball, v1 = speed of car before being hit, v2 speed of car after being hit, t1 = time it takes for the first ball to hit, t2 = time it takes for the second ball to hit, Δt = time between first and second balls thrown.

$$\because$$ t1(u-v1) = L+v1t1
The left-hand-side (LHS) here, the distance the 1st ball travels, is simply u·t1. That's because the ball is traveling at a speed u for a time t1. That will change your result for t1:
$$\therefore$$t1 = $$L/(u-2v)$$

The distance it takes the ball to catch up with the car is equal to the initial distance L plus how much the car travels during the time it takes the ball to hit the car.

(t1-Δt)(u-v1)+(t2-t1)(u-v2) = L+vt1+(t2-t1)v2
This is the equation of distance that the car travels before being hit by the second ball. The speed is different after the car being hit by the first ball since it has an additional mass.
Same error here; the 2nd snowball is traveling at speed u for a total time t2-Δt, so it's distance traveled is u·(t2-Δt).

Hint: on the RHS, the term L+v1t1 may be replaced with u·t1.

OK so here are the corrections.

$$\because$$ u·t1 = L+v1t1
$$\therefore$$ t1 = (L+ v·t1)/u

I'm still having trouble with the second equation. This is what I got.

u·(t2-Δt) = u·t1+(t2-t1)v2

The last term of this equation deals with the motion of the car after being hit by the first snow ball until being hit by the second snowball. Then we have to calculate v2, but I'll leave that for later.

Looking good so far , and I agree, leave calculation of v2 for later.

putongren said:
u·(t2-Δt) = u·t1+(t2-t1)v2
This last equation can be manipulated to determine "t2-t1" in terms of the other parameters.

## What is the "Calculus conservation of momentum problem"?

The "Calculus conservation of momentum problem" refers to a type of physics problem that involves calculating the momentum of an object or system before and after a collision or interaction. This is done using the principles of calculus, which involves finding the derivative and integral of the position and velocity functions.

## What is the equation used to solve a "Calculus conservation of momentum problem"?

The equation used to solve a "Calculus conservation of momentum problem" is the conservation of momentum equation, which states that the total momentum of a system before a collision is equal to the total momentum of the system after the collision. This equation is expressed as:
m1v1i + m2v2i = m1v1f + m2v2f
where m1 and m2 are the masses of the objects involved, and v1i, v2i, v1f, and v2f are the initial and final velocities of the objects.

## What are the steps for solving a "Calculus conservation of momentum problem"?

The steps for solving a "Calculus conservation of momentum problem" are as follows:
1. Identify the objects or systems involved in the collision.
2. Determine the initial and final velocities of the objects or systems.
3. Use the conservation of momentum equation to set up an equation with the unknown final velocities.
4. Use calculus to find the derivatives and integrals of the position and velocity functions, if necessary.
5. Solve the equation to find the final velocities of the objects or systems.
6. Check your solution by plugging in the values and ensuring that the total momentum is conserved before and after the collision.

## What are some real-life applications of "Calculus conservation of momentum problems"?

"Calculus conservation of momentum problems" have many real-life applications, such as:
- Calculating the speed and direction of a car after a collision
- Predicting the trajectory of a rocket after launch
- Understanding the motion of objects in a billiards game
- Analyzing the impact of a bullet on a target
- Investigating the movement of particles in a chemical reaction.

## What are some common mistakes when solving "Calculus conservation of momentum problems"?

Some common mistakes when solving "Calculus conservation of momentum problems" include:
- Forgetting to account for all the objects or systems involved in the collision
- Using the wrong units for mass or velocity
- Not setting up the conservation of momentum equation correctly
- Forgetting to consider external forces that may affect the momentum
- Making errors in the calculus calculations, such as forgetting to take the derivative or integral.

• Mechanics
Replies
30
Views
2K
• Mechanics
Replies
5
Views
3K
• Mechanics
Replies
40
Views
2K
• Mechanics
Replies
13
Views
2K
• Mechanics
Replies
14
Views
1K
• Mechanics
Replies
11
Views
1K
• Mechanics
Replies
43
Views
2K
• Mechanics
Replies
2
Views
1K
• Mechanics
Replies
3
Views
551
• Mechanics
Replies
3
Views
1K